## Precalculus (6th Edition) Blitzer

In order to verify the partial fraction decomposition, rational fractions are to be solved by the least common multiple and their solution is equal to the partial fraction decomposition of this solution. Let us take an example, Consider $\frac{3}{x}+\frac{5}{x+1}$ Normally, we solve this question by taking the least common multiple (L.C.M.). And the common polynomial of the denominator is $x\left( x+1 \right)$. So, \begin{align} & \frac{3}{x}+\frac{5}{x+1}=\frac{3\left( x+1 \right)+5x}{x\left( x+1 \right)} \\ & =\frac{3x+3+5x}{x\left( x+1 \right)} \\ & =\frac{8x+3}{x\left( x+1 \right)} \end{align} Then, operate the partial fraction decomposition of the above polynomial, $\frac{8x+3}{x\left( x+1 \right)}=\frac{A}{x}+\frac{B}{x+1}$ And multiply both sides by $x\left( x+1 \right)$, \begin{align} & 8x+3=A\left( x+1 \right)+Bx \\ & 8x+3=Ax+A+Bx \\ & 8x+3=\left( A+B \right)x+A \end{align} And equate the equation, \begin{align} & A=3 \\ & A+B=8 \end{align} Solving the above equation, $A=3,B=5$ Substitute these values in $\frac{8x+3}{x\left( x+1 \right)}=\frac{A}{x}+\frac{B}{x+1}$, \begin{align} & \frac{8x+3}{x\left( x+1 \right)}=\frac{3}{x}+\frac{5}{x+1} \\ & =\frac{3\left( x+1 \right)+5x}{x\left( x+1 \right)} \\ & =\frac{3x+3+5x}{x\left( x+1 \right)} \\ & =\frac{8x+3}{x\left( x+1 \right)} \end{align} Thus, in both the cases, the solution is the same. So, the result is verified.